Question

If ${\int }_0^{\pi }\frac{\sin x}{x^2}dx,then{\int }_0^{\pi /2}\frac{\cos 2x}{x}dx$ is equal to :

• A. $1-A$
• B. $3/2-A$
• C. $A-1$
• D. $1+A$

Answer 1

The correct option is D $1+A$

Given,

$A={\int }_0^{\pi }\frac{sinx}{x^2}dx$

integration by parts $u=\sin \left(x\right),v^{\prime }=\frac{1}{x^2}$

$={\left[\frac{-\sin \left(x\right)-x\cdot \int -\frac{\cos \left(x\right)}{x}dx}{x}\right]}_0^{\pi }$

$={\left[\frac{-\sin \left(x\right)-x(-\text{Ci}\left(x\right))}{x}\right]}_0^{\pi }$

$={\left[\frac{-\sin \left(x\right)+x\text{Ci}\left(x\right)}{x}\right]}_0^{\pi }$

$=\text{Ci}\left(\pi \right)+1-\text{Ci}\left(0\right)$

$A=\text{Ci}\left(\pi \right)+1-\text{Ci}\left(0\right)$

Now,

${\int }_0^{\frac{\pi }{2}}\frac{\cos \left(2x\right)}{x}dx$

substitute $u=2x$

$={\int }_0^{\pi }\frac{\cos \left(u\right)}{u}du$

$={\left[\text{Ci}\left(u\right)\right]}_0^{\pi }$

$=\text{Ci}\left(\pi \right)-\text{Ci}\left(0\right)$

$\therefore {\int }_0^{\frac{\pi }{2}}\frac{\cos \left(2x\right)}{x}dx=A+1$